Hi there!
I have purchased 2 audio books. Douglas Self blameless... and High-Power audio amplifier guidebook with 50 projects 50 to 500w...
I have been repairing amplifiers for a couple of years now and understand
most of the different parts of circuits in an basic amplifier.
Now i just realized that i dont fully understand how the "first" pre-amp input stage whatever you call it... How is this transistor biased?
I know from repairing amplifiers that you set the Base Voltage to .6 volts (just theoretical). to get rid of cross over distortion in the power stage.
This is for class AB.
What about the "input transistor".... In my head i want this to be biased like an class "A" amp because the input is AC and the voltage swing will go both positive and negative.
Look at this schematic:
www.itxpress.se/amp.jpg
If you look at Q1 the base has 0V applied to it. I can not understand the function of this circuit.
Input ac "music"... negative would draw the voltage down and positive up... basic... but if i have 0 volts already it can not go further down because there is no down.. 0 and ground is the same thing here. it can only go positive but without no current trough Q1 it would be clipping like **** at about .6 volts and up to 1 volt wich is normal line level. from .6 and down to 0 you would not hear nothing at all. and the "negative side" of the audio you would not hear at all..... of course this is not the case.....
This is some late night writing for me so dont laugh at me, am just trying to figure this out =) Can someone explain the function for me or give me some advice where to get the information?
I have purchased 2 audio books. Douglas Self blameless... and High-Power audio amplifier guidebook with 50 projects 50 to 500w...
I have been repairing amplifiers for a couple of years now and understand
most of the different parts of circuits in an basic amplifier.
Now i just realized that i dont fully understand how the "first" pre-amp input stage whatever you call it... How is this transistor biased?
I know from repairing amplifiers that you set the Base Voltage to .6 volts (just theoretical). to get rid of cross over distortion in the power stage.
This is for class AB.
What about the "input transistor".... In my head i want this to be biased like an class "A" amp because the input is AC and the voltage swing will go both positive and negative.
Look at this schematic:
www.itxpress.se/amp.jpg
If you look at Q1 the base has 0V applied to it. I can not understand the function of this circuit.
Input ac "music"... negative would draw the voltage down and positive up... basic... but if i have 0 volts already it can not go further down because there is no down.. 0 and ground is the same thing here. it can only go positive but without no current trough Q1 it would be clipping like **** at about .6 volts and up to 1 volt wich is normal line level. from .6 and down to 0 you would not hear nothing at all. and the "negative side" of the audio you would not hear at all..... of course this is not the case.....
This is some late night writing for me so dont laugh at me, am just trying to figure this out =) Can someone explain the function for me or give me some advice where to get the information?
Hi
Q1 is part of a differential amplifier formed by Q1 & Q2. The input signal is compared to the output signal (through a voltage divider) and the difference is fed to the VAS. It is true that BJT's must have some base current in order to function. If you measure the emitter voltage reference to GND of Q1 & Q2 you would see +0.6V, or the Vbe of the transistors. Base current flows through R4 to GND. The base current for Q2 flows through R10 to the output node. The bias is provided to the emitters by Q3, as a current source, it supplies the bias current for the two input devices. 0.6V is across R1, so 5mA. This 5mA is divided equally between Q1 & Q2. Q9 & Q10 form a current mirror. This circuit reflects the current on one side, the source Q10, and the reflection Q9 in attempt to keep the current in Q1 & Q2 equal. R2 and R3 provide emitter degeneration, a form of local feedback. The purpose is to reduce the percent difference of Q1 & Q2. The effect also reduces gain but trades it for bandwidth.
Q1 is part of a differential amplifier formed by Q1 & Q2. The input signal is compared to the output signal (through a voltage divider) and the difference is fed to the VAS. It is true that BJT's must have some base current in order to function. If you measure the emitter voltage reference to GND of Q1 & Q2 you would see +0.6V, or the Vbe of the transistors. Base current flows through R4 to GND. The base current for Q2 flows through R10 to the output node. The bias is provided to the emitters by Q3, as a current source, it supplies the bias current for the two input devices. 0.6V is across R1, so 5mA. This 5mA is divided equally between Q1 & Q2. Q9 & Q10 form a current mirror. This circuit reflects the current on one side, the source Q10, and the reflection Q9 in attempt to keep the current in Q1 & Q2 equal. R2 and R3 provide emitter degeneration, a form of local feedback. The purpose is to reduce the percent difference of Q1 & Q2. The effect also reduces gain but trades it for bandwidth.
Last edited:
0 volts and ground are not the same thing. The input isn't grounded, but connected to a 10k resistor(r4) which is grounded. Since it is a PNP transistor, it requires a small negative current to operate. This is the called the input bias current. This is determined by the circuit and the current gain of the transistor - if the current gain is 200 and the collector current is 2.5 ma, then the base current would be 12.5 uA (12.5 time 200 is 2500 uA or 2.5 mA.) This current is supplied by R4 from ground. So if there is 12.5 uA going through r4, there must be a voltage drop of 125 mV. Since the current is negative, the base of the input transistor is actually at +125 mV. As the signal is applied, the voltage and current at the input changes too. The circuit operates as CBS240 describes.
Last edited:
Thank you for some really good answeres!
When looking at it and reading youre posts i can se what you are telling me.
Its never good trying to solve problems late at night
Good answeres!
This must mean that there is always negative voltage on the base of
Q1 and the input (AC.. music) making the negative voltage on the base
more or less negative as the voltage changes. Is this correct?
When looking at it and reading youre posts i can se what you are telling me.
Its never good trying to solve problems late at night
Good answeres!
This must mean that there is always negative voltage on the base of
Q1 and the input (AC.. music) making the negative voltage on the base
more or less negative as the voltage changes. Is this correct?
Last edited:
You are getting there - but. The input current is negative - that means the source (ground) has to be at lower potential than the "destination" the base of q1, so q1 base is at a small positive voltage. The AC signal will cause that voltage to go up an down, offset by that amount - a 1 volt peak to peak sinewave in, at the base will be going from -.375 to +.625 If the operating point (or bias or input offset) is +.125 V. The voltages at the emitter and the base of Q2 will also go up and down around their operating voltage - is the emitter is at +.725 volts, it will go from +.225 to +1.225, and the base of Q2 will have the same signal as Q1. The most difficult thing to visualize is that because of feed back, the actual changes of voltages are very small for a very large change in output - typical amp has a gain of 10000 to 100000 before feed back is applied. Keep asking questions and eventually the big picture will become clearer. Steve
Last edited:
The DC may be +0.125V but the AC input on the base of Q1 follows the input signal, as does the emitter. The base of Q1 is always negative 0.6V from the emitter of Q1 because it is a PNP BJT. The base of Q2 is driven by the output via the feedback voltage divider to have the exact equal signal as the input to Q1, ideally. Thus the emitter of Q2 follows the signal as well. The difference of those two signals, the source input and the feedback input, is fed to the VAS to complete the control loop.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.